You have a piece of aluminium whose mass is
8
0
0
g
, and suppose it is heated up to
1
0
0
0
∘
C
.
Now, given that you are cooling it down to 1 5 0 ∘ C , calculate the amount of heat (in Joules) given out during the process of cooling.
Details and assumptions:
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My solution is almost similar to that of Sravanth but I like celsius more than kelvin.
C a l u m i n i u m = 9 0 0 J / k g K = 9 0 0 J / k g C because a unit change in celsius is equal to unit change in kelvin.So the formula for release of heat is given by :
Δ Q = m . C a l u m i n i u m Δ T = ( 0 . 8 ) ( 9 0 0 ) ( 1 0 0 0 − 1 5 0 ) = 6 1 2 0 0 0 J
Oh yeah! The change in temperatures in both the scales is the same.
C H E E R S ! ! !
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Have some room for me back there guys? xD
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The following values are given,
m a s s = 8 0 0 g = 0 . 8 k g I n i t i a l t e m p e r a t u r e = T 1 = 1 0 0 0 ° C = 1 2 7 3 ° K F i n a l t e m p e r a t u r e = 1 5 0 ° C = 4 2 3 ° K C a l u m i n i u m = 9 0 0 J / k g K
We know that heat energy Q is given by the following expression,
Q = C m Δ T = 9 0 0 × 0 . 8 × ( 1 2 7 3 − 4 2 3 ) = 7 2 0 × 8 5 0 = 6 1 2 0 0 0 J